What is the general solution for sqrt(7)sin(θ)-1=0 ?

The general solution for sqrt(7)sin(θ)-1=0 is θ=0.38759…+2pin,θ=pi-0.38759…+2pin

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sqrt(7)sin(θ)-1=0

7 sin (θ) - 1 = 0

θ = 0.38759 \dots + 2 πn, θ = π - 0.38759 \dots + 2 πn

Degrees

θ = 22.20765 \dots^{\circ} + 36 0^{\circ} n, θ = 157.79234 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 sin (θ) - 1 = 0

Move

1

to the right side

7 sin (θ) - 1 = 0

Add

1

to both sides

7 sin (θ) - 1 + 1 = 0 + 1

Simplify

7 sin (θ) = 1

7 sin (θ) = 1

Divide both sides by

7

7 sin (θ) = 1

Divide both sides by

7

\frac{7 sin ( θ )}{7} = \frac{1}{7}

Simplify

\frac{7 sin ( θ )}{7} = \frac{1}{7}

Simplify

\frac{7 sin ( θ )}{7} : sin (θ)

\frac{7 sin ( θ )}{7}

Cancel the common factor:

7

= sin (θ)

Simplify

\frac{1}{7} : \frac{7}{7}

\frac{1}{7}

Multiply by the conjugate

\frac{7}{7}

= \frac{1 \cdot 7}{7 7}

1 \cdot 7 = 7

77 = 7

77

Apply radical rule:

a a = a

77 = 7

= 7

= \frac{7}{7}

sin (θ) = \frac{7}{7}

sin (θ) = \frac{7}{7}

sin (θ) = \frac{7}{7}

Apply trig inverse properties

sin (θ) = \frac{7}{7}

General solutions for

sin (θ) = \frac{7}{7}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{7}{7}) + 2 πn, θ = π - arcsin (\frac{7}{7}) + 2 πn

θ = arcsin (\frac{7}{7}) + 2 πn, θ = π - arcsin (\frac{7}{7}) + 2 πn

Show solutions in decimal form

θ = 0.38759 \dots + 2 πn, θ = π - 0.38759 \dots + 2 πn

sin (3 θ) = - \frac{1}{2}

2 cos (θ) = 2

2 sin (x) - cot (x) - csc (x) = 0

tan (θ) = \frac{- 4}{2}

cos (x - \frac{π}{6}) = 0, 5

What is the general solution for sqrt(7)sin(θ)-1=0 ?
The general solution for sqrt(7)sin(θ)-1=0 is θ=0.38759…+2pin,θ=pi-0.38759…+2pin